1 Answer
1

If you interchange the roles of $x$ and $x+h$ (and replace $h$ by $-h$), then you see that
$$
|f(x)-f(x+h)+f'(x+h)(h)|<\epsilon|h|,
$$
which, upon combining with what you have, and using the triangle inequality, shows that $f'(x)$ and $f'(x+h)$ are $\epsilon$-close.